Prasanna <prasannaprakash <at> gmail.com> writes: > > Dear R experts > > I am looking for a package which gives me latin hyper cube samples > from the grid of values produced from the command "expand.grid". Any > pointers to this issue might be very useful. Basically, I am doing the > following: > > > a<-(1:10) > > b<-(20:30) > > dataGrid<-expand.grid(a,b) > > Now, is there a way to use this "dataGrid" in the package "lhs" to get > latin hyper cube samples. > > Thanking you > Prasanna > > ______________________________________________ > R-help <at> stat.math.ethz.ch mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. > >
Prasanna, I think I understand your question, please let me know if this explanation is not what you need. Since, lhs is a contributed package, you could contact me directly first. a <- 1:10 b <- 20:30 dataGrid <- expand.grid(a, b) I believe that you want a Latin hypercube sample from the integers 1-10 in the first variable and 20-30 in the second. I will offer a way to do something similar with the lhs package, but then also offer alternatives way which may meet your needs better. The lhs package returns a uniformly distributed stratified sample from the unit hypercube. The marginal distributions can then be transformed to your distribution of choice. If you wanted a uniform Latin hypercube on [1,10] and [20,30] with 22 samples, you could do: require(lhs) X <- randomLHS(22, 2) X[,1] <- 1+9*X[,1] X[,2] <- 20+10*X[,2] X OR X <- randomLHS(22, 2) X[,1] <- qunif(X[,1], 1, 9) X[,2] <- qunif(X[,2], 20, 30) X Since I think you want integers (which I haven't thought about before now), then I think we must be careful about what we mean by a Latin hypercube sample. If you wanted exactly 3 points, then you could divide up the range [1,10] into three almost equal parts and sample from 1:3, 4:6, and 7:10. The problem is that it wouldn't be uniform sample across the range. (7 would be sampled less often than 2 for example) I think that to do a Latin hypercube sample on the intgers, you should have a number of integers on the margins which have the number of points sampled as a common factor. For example if you sample 3 points from 1:9, and 21:32 then you could sample as follows: a <- c(sample(1:3,1), sample(4:6, 1), sample(7:9, 1)) b <- c(sample(21:24,1), sample(25:28, 1), sample(29:32,1)) and then randomly permute the entries of a and b. Or more generally, take n samples from the list of integer groups: integerLHS <- function(n, intGroups) { stopifnot(all(lapply(intGroups, function(X) length(X)%%n)==0)) stopifnot(require(lhs)) stopifnot(is.list(intGroups)) ranges <- lapply(intGroups, function(X) max(X)-min(X)) A <- matrix(nrow=n, ncol=length(intGroups)) for(j in 1:length(ranges)) { sequ <- order(runif(n)) if(length(intGroups[[1]]) > 1) { spacing <- intGroups[[j]][2]-intGroups[[j]][1] } else stop("must have more than 1 intGroup") for(k in 1:n) { i <- sequ[k] a <- min(intGroups[[j]])+(i-1)*(ranges[[j]]+spacing)/n b <- min(intGroups[[j]])+i*(ranges[[j]]+spacing)/n-1 if(a < b) { A[k,j] <- sample(seq(a,b,spacing), 1) } else if(a==b) { A[k,j] <- a } else stop("error") } } return(A) } integerLHS(10, list(1:10, 31:40)) integerLHS(5, list(1:10, 31:40)) integerLHS(2, list(1:10, 31:40)) integerLHS(5, list(1:20, 31:60, 101:115)) integerLHS(5, list(seq(2,20,2), 31:60, 101:115)) The function above is neither efficient nor tested, but it is a place for you to start. Rob ______________________________________________ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.